1.题目链接。给定的一张图，有三个人A，B，C，和两个集合s1，s2.A会等概率的在集合s1中的点出现，B会等概率的在集合s2中的点出现，C会等概率的出现在图上的任意一个点。现在A,B,C要在图中的某个点集合，问他们所走过的最短路程的期望是多少？

2.在图中新建一个点(n+1),然后定义个点到这个点的距离就是A，B所在点到这个点距离最短距离之和。然后C点到这个点的每一条最短路就对答案有贡献。所以枚举s1,s2集合，对于每一种情况，新建图之后跑最短路。但是这里很特殊的一点是，每个边的边权都是1，那么可以用BFS+双队列优化掉dij的log，这样就可以A掉该题。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll __gcd(ll a, ll b)
{
  return !b ? a : __gcd(b, a % b);
}

struct dijnode {
  int d, u;
  dijnode(int d = 0, int u = 0) :d(d), u(u) {}
};

struct graph {
  static const int maxn = 1e5 + 5;
  static const int maxm = 2e5 + 5;
  struct star {
    int v, w, nex;
    star(int v = 0, int w = 0, int nex = 0) :v(v), w(w), nex(nex) {}
  }edge[maxm];
  int head[maxn], tot, n;
  int d[maxn];

  void init(int _n) {
    n = _n;
    tot = -1;
    memset(head, -1, (n + 1) * sizeof(head[0]));
  }

  void add_edge(int u, int v, int w) {
    edge[++tot] = star(v, w, head[u]);
    head[u] = tot;
  }
  void short_path(queue<dijnode>& q2, int s, int* dist) {//short path
    queue<dijnode>q1;// dis and vert
    dist[s] = 0;
    q1.push(dijnode(dist[s], s));
    while (!q1.empty() || !q2.empty())
    {
      dijnode nd;
      if (q1.empty()) nd = q2.front(), q2.pop();
      else if (q2.empty()) nd = q1.front(), q1.pop();
      else if (q1.front().d < q2.front().d) nd = q1.front(), q1.pop();
      else nd = q2.front(), q2.pop();
      if (nd.d != dist[nd.u])continue;
      int u = nd.u;

      for (int i = head[u]; ~i; i = edge[i].nex)
      {
        int v = edge[i].v, w = edge[i].w;
        if (dist[u] + w < dist[v]) {
          dist[v] = dist[u] + w;
          q1.push(dijnode(dist[v], v));
        }
      }
    }
  }
}g1;

int n, m;
const int maxn = 1e5 + 55;
int d1[20][maxn], d2[20][maxn];

vector<int> vec[maxn];

long long solve(int p1, int p2) {
  queue<dijnode> q;
  int mx = 0;
  /*
    vec数组保存的时 vec[x]=i.距离为x的点有哪些？？？
  */
  for (int i = 1; i <= n; i++)
  {
    vec[d1[p1][i] + d2[p2][i]].push_back(i);
    g1.d[i] = d1[p1][i] + d2[p2][i];
    mx = max(mx, g1.d[i]);
  }
  for (int i = 0; i <= mx; i++) {
    for (int x : vec[i]) q.push(dijnode{ d1[p1][x] + d2[p2][x],x });
  }
  for (int i = 1; i <= n; i++) vec[d1[p1][i] + d2[p2][i]].clear();
  g1.short_path(q, n + 1, g1.d);
  long long sum = 0;
  for (int i = 1; i <= n; i++) sum += g1.d[i];
  return sum;
}



int main() {
  int t;
  scanf("%d", &t);
  for (int ti = 1; ti <= t; ti++) {
    scanf("%d%d", &n, &m);
    g1.init(n + 1);
    for (int i = 0; i < m; i++) {
      int u, v; scanf("%d%d", &u, &v);
      g1.add_edge(u, v, 1);
      g1.add_edge(v, u, 1);
    }
    int s1; scanf("%d", &s1);
    for (int i = 0; i < s1; i++)
    {
      int v1;
      scanf("%d", &v1);
      queue<dijnode> q;
      for (int j = 1; j <= n; j++) d1[i][j] = 2e9;
      //求出v1到 1-n的最短路，保存在d1[i]这个数数组里
      g1.short_path(q, v1, d1[i]);
    }

    int s2; scanf("%d", &s2);
    for (int i = 0; i < s2; i++) {
      int v2; scanf("%d", &v2);
      queue<dijnode> q;
      for (int j = 1; j <= n; j++) d2[i][j] = 2e9;
      g1.short_path(q, v2, d2[i]);
    }

    long long sum = 0;
    for (int i = 0; i < s1; i++)
    {
      for (int j = 0; j < s2; j++)
      {
        sum += solve(i, j);
      }
    }

    long long fenmu = 1ll * s1 * s2 * n;
    long long gcd = __gcd(sum, fenmu);
    printf("Case #%d: %lld/%lld\n", ti, sum / gcd, fenmu / gcd);
  }
}